WAEC 2023 · Paper 2 · Q14

  1. (a)

    A particle is projected at an angle of 42∘42^\circ with a velocity of 68 m s−168\text{ m s}^{-1}. Calculate, correct to one decimal place, the: (i) greatest height travelled; (ii) time of flight; (iii) horizontal range travelled. [g=10 m s−2][g = 10\text{ m s}^{-2}] Enter the three values.

    Separate values with commas, e.g. 3, −2

  2. (b)

    A non-uniform beam PQPQ of length 15 m15\text{ m} and weight 120 N120\text{ N} rests horizontally on supports at PP and QQ. If the centre of gravity of the beam is 3.5 m3.5\text{ m} from QQ, find the reaction at QQ.

Worked solution (try it first)

(a)(i)

  1. H=u2sin⁡2θ2gH = \dfrac{u^2\sin^2\theta}{2g}
    =682sin⁡242∘20= \dfrac{68^2\sin^2 42^\circ}{20}
    ≈103.5 m\approx 103.5\text{ m}.

(ii)

  1. T=2usin⁡θgT = \dfrac{2u\sin\theta}{g}
    =136sin⁡42∘10= \dfrac{136\sin42^\circ}{10}
    ≈9.1 s\approx 9.1\text{ s}.

(iii)

  1. R=u2sin⁡2θgR = \dfrac{u^2\sin2\theta}{g}
    =4624sin⁡84∘10= \dfrac{4624\sin84^\circ}{10}
    ≈459.9 m\approx 459.9\text{ m}.

(b)

  1. The weight acts at the centre of gravity, 15−3.5=11.5 m15 - 3.5 = 11.5\text{ m} from PP.
  2. Moments about PP: 15RQ=120×11.5=138015R_Q = 120 \times 11.5 = 1380, so RQ=92 NR_Q = 92\text{ N}.

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